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Ex 6.4, 3 - ABC and DBC are two triangles on same base BC - Area of similar triangles

Ex 6.4, 3 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.4, 3 - Chapter 6 Class 10 Triangles - Part 3
Ex 6.4, 3 - Chapter 6 Class 10 Triangles - Part 4

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Ex 6.4, 3 In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (π‘Žπ‘Ÿ(𝐴𝐡𝐢))/(π‘Žπ‘Ÿ(𝐷𝐡𝐢)) = 𝐴𝑂/𝐷𝑂 Given: βˆ† ABC and βˆ† DBC Having common base BC To prove: (π‘Žπ‘Ÿ βˆ† 𝐴𝐡𝐢)/(π‘Žπ‘Ÿ βˆ† 𝐷𝐡𝐢)=𝐴𝑂/𝐷𝑂 Proof: Since AO and OD are not part of βˆ† 𝐴𝐡𝐢 and βˆ† 𝐷𝐡𝐢 , we cannot directly use the theorem We know that Area of triangle = 1/2Γ—π΅π‘Žπ‘ π‘’Γ—π΄π‘™π‘‘π‘–π‘‘π‘’π‘‘π‘’ Lets draw altitude AE [ AE βŠ₯𝐡𝐢 ] Hence, ar βˆ† ABC=1/2×𝐡𝐢×𝐴𝐸 Similarly, for Ξ” DBC Let us draw altitude DF (DF βŠ₯ BC) Hence, ar βˆ† DBC=1/2×𝐡𝐢×𝐷𝐹 Now, taking ratio (π‘Žπ‘Ÿ βˆ†π΄π΅πΆ)/(π‘Žπ‘Ÿ βˆ†π·π΅πΆ) = (1/2 Γ— 𝐡𝐢 Γ— 𝐴𝐸)/(1/2 Γ— 𝐡𝐢 Γ— 𝐷𝐹) (π‘Žπ‘Ÿ βˆ†π΄π΅πΆ)/(π‘Žπ‘Ÿ βˆ†π·π΅πΆ) = 𝐴𝐸/𝐷𝐹 Now in βˆ† AOE and Ξ” DOF ∠ AEO = ∠ DFO ∠ AOE = ∠ DOF So, by using AA similarity criterion βˆ† AOE ~ βˆ† DOF If two triangle are similar their , corresponding sides are in the same ratio So, 𝐴𝐸/𝐷𝐹=𝐴𝑂/𝐷𝑂 Putting (2) in (1) (π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐴𝐡𝐢)/(π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐷𝐡𝐢)=𝐴𝐸/𝐷𝐹 (π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐴𝐡𝐢)/(π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐷𝐡𝐢)=𝐴𝑂/𝐷𝑂 Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.